Question

Determine the value of .

integration pi to -pi.
2x(1+sinx)/1+cos^2x. × dx

Answer 1

Heya
§(this will be sign of integration )


(-π to π) § 2x(1+sinx)/1+ cos²x. × dx

= (-π to π) § 2x/1+cos²x × dx + (π to -π) § 2xsinx/1+cos²x × dx
= Now let the first integral as A and second as B

• Now it is clear that A = f(x) is odd function
since f(x) = -f(x) .
so § A = 0

•Now we will check for B
and this is clear an even function.


B = (-π to π) § 2xsinx/1+cos²x ×dx = 2.2[0 to π]§xsinx/1+cos²x. × dx.

= 4 (0 to π) § (π-x)sin(π-x)/1+cos²(π-x) ×dx
=4π§(0toπ) sinx/1+cos²x × dx. - B
so
2B = 4π§(0toπ) sinx/1+cos²x ×dx
plug cosx= t
so -sinxdx => dt
plug this value here

2B = 4π§(0to1) dt/1+t²
so => 8π[tan-¹t](0to1) = 8π(tan -¹ 1 - tan-¹0)
=> 8π( π/4) => 2π²
now 2B = 2π² => B = π²
now A + B = 0+π²
so
(-π to π) § 2x(1+sinx)/1+ cos²x. × dx = π².


Any doubts regarding this pls comment.